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Question: Find the value of \(\tan \left( {\dfrac{{5\pi }}{3}} \right)\)....

Find the value of tan⁡(5π3)\tan \left( {\dfrac{{5\pi }}{3}} \right).

Explanation

Solution

The Cartesian system divides the plane into 4 different quadrants.
Quadrant I: 0  −  π20\; - \;\dfrac{\pi }{2}
Quadrant II: π2  −  π\dfrac{\pi }{2}\; - \;\pi
Quadrant III: π  −  3π2\pi \; - \;\dfrac{{3\pi }}{2}
Quadrant IV: 3π2  −  2π\dfrac{{3\pi }}{2}\; - \;2\pi
So in order to find the value of tan⁡(5π3)\tan \left( {\dfrac{{5\pi }}{3}} \right) we first must convert (5π3)\left( {\dfrac{{5\pi }}{3}} \right)as the sum of two numbers such that one of them forms the boundary of any quadrant and the other value acts as a reference value whose tan value is known to us.

Complete step by step solution:
Given
tan⁡(5π3)  ....................(i)\tan \left( {\dfrac{{5\pi }}{3}} \right)\;....................\left( i \right)
We know that we have to convert (5π3)\left( {\dfrac{{5\pi }}{3}} \right)as the sum of two numbers as mentioned above. So converting(5π3)\left( {\dfrac{{5\pi }}{3}} \right):
(5π3)=(π  +  2π3)  ...............(ii)\left( {\dfrac{{5\pi }}{3}} \right) = \left( {\pi \; + \;\dfrac{{2\pi }}{3}} \right)\;...............\left( {ii} \right)
Here π\pi is a boundary of the quadrant also we can find the value of tan⁡(2π3)\tan \left( {\dfrac{{2\pi }}{3}} \right) ⇒tan⁡(5π3)  =  tan⁡(π+2π3)=tan⁡(2π3) \Rightarrow \tan \left( {\dfrac{{5\pi }}{3}} \right)\; = \;\tan \left( {\pi + \dfrac{{2\pi }}{3}} \right) = \tan \left( {\dfrac{{2\pi }}{3}} \right)
Now converting (2π3)\left( {\dfrac{{2\pi }}{3}} \right) in the same manner as that of (5π3)\left( {\dfrac{{5\pi }}{3}} \right).
⇒(2π3)  =  (π−π3).....................(iii)\Rightarrow \left( {\dfrac{{2\pi }}{3}} \right)\; = \;\left( {\pi - \dfrac{\pi }{3}} \right).....................\left( {iii} \right)
So here π\pi is a boundary of the quadrant, also know the value of tan⁡(π3)\tan \left( {\dfrac{\pi }{3}} \right)
⇒tan⁡(2π3)  =tan⁡(π−π3)  = −tan⁡(π3).............(iv)\Rightarrow \tan \left( {\dfrac{{2\pi }}{3}} \right)\; = \tan \left( {\pi - \dfrac{\pi }{3}} \right)\; = \, - \tan \left( {\dfrac{\pi }{3}} \right).............\left( {iv} \right)
Here the negative sign occurs since (π−π3)\left( {\pi - \dfrac{\pi }{3}} \right) is the Quadrant II and tan is negative in the Quadrant II.
So:

1.732050808...........\left( v \right)$$ Hence our final answer is given in $\left( v \right)$which is $ - \sqrt 3 = \; - 1.732050808$ **Therefore: $\tan \left( {\dfrac{{5\pi }}{3}} \right)\; = - \sqrt 3 = \; - 1.732050808$** **Note:** tan is the abbreviation of tangent. Also: Quadrant I: $0\; - \;\dfrac{\pi }{2}$ All values are positive. Quadrant II: $\dfrac{\pi }{2}\; - \;\pi $ Only Sine and Cosec values are positive. Quadrant III: $\pi \; - \;\dfrac{{3\pi }}{2}$ Only Tan and Cot values are positive. Quadrant IV :$\dfrac{{3\pi }}{2}\; - \;2\pi $ Only Cos and Sec values are positive. Also while approaching a trigonometric problem one should keep in mind that one should work with one side at a time and manipulate it to the other side. The most straightforward way to do this is to simplify one side to the other directly, but we can also transform both sides to a common expression if we see no direct way to connect the two.